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Ln of a negative numer - please help!

  1. Nov 14, 2004 #1
    Hey there, I'm trying to work out all the solutions of z for ln(z)=-1. I let -1=ln|z| so then I took exponentials of both sides so I had e^(-1)=|z| so 1/e would equal z. I wasn't sure about having z also equal to -1/e, because since ln(z) = ln|z| + iarg(z) then for z=-1/e then the arg (z) would have to be pi or something..... If anyone could help it would be great cos with the way the question has been asked about finding ALL the solutions, I think I'm missing something here. Thanks. :smile:
     
  2. jcsd
  3. Nov 14, 2004 #2

    Integral

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    In my text,

    log(z) = log r + i [itex] \Theta [/itex]

    r = |z|

    [itex] \Theta [/itex]= arg z
     
  4. Nov 14, 2004 #3

    Gokul43201

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    As suggested above, simply write [itex]z=re^{i\theta} [/itex] and compare real and imaginary parts.
     
  5. Nov 15, 2004 #4
    Yeah, I did that but I just got 1/e.... I'm getting kinda confused. Could you have something like (1/e)e^(2mpi) where m is 0, +1, -1 etc...?
     
  6. Nov 15, 2004 #5

    dextercioby

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    No,you're left just with the "ugly" solution [tex] z=e^{-&1} [/tex].
    Remember,even for imaginary arguments,the exponential and the natural logarithm are still one the inverse of the other.
     
  7. Nov 15, 2004 #6
    That's all I had to do? So I definitely don't have to have different values for the argument like 0 (which was what I thought we had here) and then +2pi, -2pi etc? Cos I wasn't sure if by just putting down e^-1 I was just calculating the principal value.....
     
  8. Nov 15, 2004 #7

    Integral

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    Did you look at and understand mine and Gokuls posts?

    for ln(-1)

    r = |-1| = 1

    [itex] \Theta= \pi [/itex]

    So the PRINCIPLE Value of Ln(-1) = ln 1 + i [itex] \pi [/itex]

    With a bit of thought you ought to be able to find the rest of the solutions.

    Edit: If you have a real number answer there must be something wrong, since ln (-1) does not exist in the Reals. Isn't it obvious that your answer must be imaginary?
     
    Last edited: Nov 15, 2004
  9. Nov 15, 2004 #8
    I'm confused (story of my life). I'm finding ln(z) = -1, so ln|z| +iarg(z) =-1, so I don't really get where the ln(-1) bit comes from.... I just thought you'd compare real and imaginary parts or something here.

    Thanks everyone for helping me btw, I really appreciate it. :smile:
     
  10. Nov 15, 2004 #9
    But my z is real is it not, since it's 1/e or something....
     
  11. Nov 15, 2004 #10

    learningphysics

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    Claire, I believe you have the right answer with z=1/e. Don't think there's any other solution.
     
  12. Nov 15, 2004 #11

    learningphysics

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    I apologize for my previous post!

    >>Yeah, I did that but I just got 1/e.... I'm getting kinda confused. Could you have something like (1/e)e^(2mpi) where m is 0, +1, -1 etc...?<<

    Yes.

    Instead of using your previous formula (which gives only principal value) use the generalized one
    ln(z) = ln|z| + i ([tex]\theta[/tex]+2*k*pi) where k is any integer.

    Since the i component needs to be zero, solve for [tex]\theta[/tex].
     
  13. Nov 15, 2004 #12
    S would we just have -1 = ln|z| + iarg(z) so the ln|z| part would give us the real part (say r), which would be 1/e. Then arg (z)= 2kpi where k is any integer. But would you then have to write this in polar form to indicate ALL the solutions cos writing it in cartesian form wouldn't give this, right? So would we have to write it like re^(itheta)? Would that then give us (1/e)e^(i2kpi) which would the give us e^(-1 + i2kpi)? Or would you not need to bother with that very last bit there? Or would it be better to write it in the form z = r(costheta + isintheta)?
     
  14. Nov 15, 2004 #13

    learningphysics

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    It would be -1=ln|z|+i(arg(z)+2kpi)

    arg(z)+2kpi=0 => arg(z)=-2kpi

    since k is any integer we can rewrite, arg(z)=2kpi and get rid of the '-'.

    so you'd have z=(1/e)e^(i2kpi) The thing is that e^(i2kpi)=1 for any integer k, so I'm not sure if you should give (1) z=1/e, or (2) z=(1/e)e^(i2kpi) as your final answer.

    They are technically equivalent but the second shows that you considered all values of arg(z), whereas the first is more simplified.

    Hope this helps.
     
  15. Nov 15, 2004 #14
    Thanks for that. :smile:

    Sorry if I'm being super-dozey here, but why is e^(i2kpi) = 1 for any integer k?
     
  16. Nov 15, 2004 #15
    Btw, I get that it would be for k=0, but not for the other k values.......
     
  17. Nov 15, 2004 #16

    StatusX

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    there will only be one solution, 1/e. ln(x) is a multivalued function in the complex plane, but its inverse e^x isn't. To solve this equation, you just apply the inverse to both sides, and since its single valued, you get a single answer. An analogy would be sin and arcsin. the equation arcsin(z)=0.5 has only one solution, z=sin(0.5), but the equation sin(z)=0.5 has an infinite number of solutions, z=arcsin(0.5)+2*pi*n for all integers n.
     
  18. Nov 15, 2004 #17
    I think I like Status's answer, as it more appeals to my mathematical common sense.
     
  19. Nov 15, 2004 #18

    learningphysics

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    e^(i2kpi)=cos(2kpi) + i sin (2kpi)=1+i0=1 (since cos 2kpi is always 1, and sin2kpi is always 0 for any integer k)

    Claire, look at Status' post. I think he gives the easiest solution.
     
  20. Nov 15, 2004 #19

    Integral

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    My apologies, I, for some reason was reading z = Ln(-1), which is a bit more interesting question of complex analysis.
     
  21. Nov 16, 2004 #20
    No problem. Thanks everyone for all your help. I was even asking my lecurer about it today and he wasn't sure so you've been great. Thank you. :smile:
     
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